Problem: Multiply the following rational expressions and simplify the result. $\dfrac{x^6+4x^3+4}{5x^2-20x}\cdot\dfrac{5x^2-10x}{x^6+6x^3+8}=$
Solution: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator of the first expression $x^6+4x^3+4$ can be factored as $(x^3+2)(x^3+2)$ using the perfect square pattern. The denominator of the first expression $5x^2-20x$ can be factored as $5x(x-4)$ by factoring out $5x$. The numerator of the second expression $5x^2-10x$ can be factored as $5x(x-2)$ by factoring out $5x$. The denominator of the second expression $x^6+6x^3+8$ can be factored as $(x^3+4)(x^3+2)$ using the sum-product pattern. Now the product looks as follows: $ \dfrac{(x^3+2)(x^3+2)}{5x(x-4)}\cdot\dfrac{5x(x-2)}{(x^3+4)(x^3+2)}$ To find the product of two rational expressions, we multiply across, then simplify: [What's that?] $\phantom{=}\dfrac{(x^3+2)(x^3+2)}{5x(x-4)}\cdot\dfrac{5x(x-2)}{(x^3+4)(x^3+2)}$ $\begin{aligned}&=\dfrac{(x^3+2)(x^3+2)\cdot 5x(x-2)}{5x(x-4)\cdot (x^3+4)(x^3+2)}&\text{Multiply across.}\\\\\\ &=\dfrac{\cancel{{(x^3+2)}}(x^3+2)\cancel{{(5x)}}(x-2)}{\cancel{{5x}}(x-4)(x^3+4)\cancel{{(x^3+2)}}}&\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{(x^3+2)(x-2)}{(x^3+4)(x-4)}\end{aligned}$ Therefore, the simplified form of the product is $\dfrac{(x^3+2)(x-2)}{(x^3+4)(x-4)}$.